One day a leetcode #342. Power of Four__github

One leetcode a day .

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Given an integer (Signed/bits), write a function to check whether it are a power of 4.

Example:
Given num =, return true. Given num = 5, return false.

Follow up:could You solve it without loops/recursion? (ii) problem solving

The main idea: to determine whether a number is not 4 of the N-second square number.
Problem-solving thinking: can refer to: "One day a Leetcode" #326. Power of three and "one day Leetcode" #231. Power of two
First, the easiest way to do this is by looping:

Class Solution {public
:
    bool ispoweroffour (int num) {
        int n = num;
        while (n>0&&n%4==0)
        {
            n/=4
        }
        return n==1;
    }
;

There are questions in the question that can not do without circulation and return to do, and judge 3 of the N-second square number, 4 of the number of n times in the plastic number of only 16, can quickly enumerate, also is a good way.
In addition, we can see that the N-second square number of 4 and the N-second square number of 2 are able to get the n-th number of integers of 4 in 2 of the n-th square number.

Class Solution {public
:
    //Three conditions: greater than 0, is 2 of the N-second square number, after the root of the integer
    bool ispoweroffour (int num) {return
        (num>0) &&! (num& (num-1)) && (sqrt (num) *sqrt (num) ==num);
    }
;

It was ingenious to see such an answer in Leetcode's discussion area.
First determine whether NUM is 2 N-second square number, in judging him and the 0xaaaaaaaa phase is 0.
A characteristic of the N-second square number of 4 is that at bit bit, the 0,2,4,6,8,10 is 1,
So, first judge is not 3 of the N-second square number, it is guaranteed to have and only one on the number of 1, and then eliminate the odd digits of 1 of the number, the rest is 4 of the N-second square number.

Class Solution {public
:
    bool ispoweroffour (int num) {
        //three conditions: is greater than 0, is 2 of the N-second square number, and has only even digits on 1 return
        (num> 0) &&! (num& (num-1)) &&! (NUM&0XAAAAAAAA);
    }
;